For more information, please see full course syllabus of Linear Algebra

For more information, please see full course syllabus of Linear Algebra

### Similar Matrices & Diagonalization

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- Intro 0:00
- Similar Matrices and Diagonalization 0:25
- Definition 1
- Example 1
- Properties
- Definition 2
- Theorem 1
- Example 3
- Theorem 2
- Example 4
- Example 5
- Procedure for Diagonalizing Matrix A: Step 1
- Procedure for Diagonalizing Matrix A: Step 2
- Procedure for Diagonalizing Matrix A: Step 3
- Procedure for Diagonalizing Matrix A: Step 4

### Linear Algebra Online Course

### Transcription: Similar Matrices & Diagonalization

*Welcome back to Educator.com and welcome back to linear algebra.*0000

*In our previous lesson, we introduced the notion of Eigenvector and Eigenvalue.*0004

*Again, very, very profoundly profoundly important concepts throughout mathematics and science.*0009

*Today, we are going to dig a little bit deeper and we are going to introduce the notion of similar matrices and the idea of diagonalization.*0015

*So, let us jump right on in. Let us start with a definition... Let me go to blue ink here.*0025

*Okay. A matrix b is said to be similar to matrix a if there is a non-singular matrix, p, such that... let us see what this definition says.*0036

*If I have some matrix a, and I find some other matrix p, and if I multiply on the left of a by p inverse, and on the right by p, so if I take p inverse a × p, the matrix that I get, b, I say that b is similar to a.*0083

*So, there is a relationship that exists between, if I can actually sandwich this matrix a between some matrix p and the inverse of p.*0102

*In the course of this lesson, we are going to talk to you actually about how to find this matrix p, and about what this matrix b looks like. Really, really quite beautiful.*0109

*Okay. A quick example just so you see what this looks like in real life. So, if I let the matrix a equal (1,1,-2,4), just a little 2 by 2... and if I say I had p which is (1,1,1,2), well, if I calculate the inverse of p, that is going to equal (2,-1,-1,1)... okay?*0120

*Now, as it turns out, if I take b, if I actually do p inverse × a × p so I multiply that by that, then by that, I end up with the following... I end up with the matrix (2,0,0,3).*0149

*Now, what you might want to do is take a look at the previous lesson. This matrix a, you remember, and I will discuss this in a minute... just to let you know what is coming up, what is coming ahead, this matrix a, we have dealt with this matrix before in the previous lesson.*0170

*We found the Eigenvalues for it. The Eigenvalues were 2 and 3. Well, notice what we did.*0183

*We found this matrix p... and we took the inverse of that, we multiplied p inverse a p, and we end up with a diagonal matrix, where the entries on the diagonal are exactly the Eigenvalues of a. That is what is going to end up being so beautiful.*0189

*Here is what is even better. If you remember the two Eigenvectors that we found for the two Eigenvalues 2 and 3 were exactly of the form (1,1,1,2).*0203

*So, this matrix p is going to end up being made up of the actual Eigenvectors for the Eigenvalues. We will just... a little preview of what is coming.*0212

*Okay. Just some quick properties of similarity. So, the first property is a is similar to a, of course... intuitively clear.*0223

*If b is similar to a, then a is similar to b. That just means that... we can always multiply on the left by p, here, and p × p inverse, this goes away, and we multiply on the right by p inverse so this is just the same.*0244

*And three... If a is similar to b, and b is similar to c, then, by transitivity, c is similar to a... I am sorry, a is similar to c, which means that c is similar to a by property too... then a is similar to c.*0262

*So, standard properties... we will be using those in a second. Another definition. We say matrix a is diagonalizable... did I spell that correct?... diagonalizable... if it is similar to a diagonal matrix.*0293

*In this case, we say a can be diagonalized -- put a comma there, in this case we say... a can be diagonalized.*0344

*Okay. Now, let us see what we have got. Alright. Profoundly, profoundly, profoundly important theorem.*0367

*An n by n matrix is diagonalizable if, and only if, it has n linearly independent Eigenvectors.*0387

*In this case, a is similar to a diagonal matrix d, where d is equal to p inverse a... p... and the diagonal elements of p, whose diagonal elements are the Eigenvalues of a.*0423

*It is like we did before. It is similar to the diagonal matrix b, and the entries on that diagonal are precisely the Eigenvalues of a.*0468

*Well, p is the matrix whose columns respectively... it means respective to the individual Eigenvalues... so Eigenvalue Λ1 gives column 1, Λ2 gives column 2, Λ3 gives column 3... so on and so forth, respectively.*0481

*Each of these columns, respectively, are the n linearly independent Eigenvectors of a.*0512

*So, again, this is what we did in our example. So now we state it as a theorem... an n by n matrix is diagonalizable if and only if it has n linearly independent Eigenvectors.*0529

*So, if I have a 3 by 3, I need 3 Eigenvectors. If I have a 5 by 5, I need 5 linearly independent Eigenvectors.*0542

*In this case, the a is similar to a diagonal matrix d, whose diagonal elements are precisely the Eigenvalues of a, and the matrix p in this relation here, is the matrix made up of the columns that are the n independent, linearly independent Eigenvectors of a.*0548

*So, from a, I can derive the matrix d, I can derive the matrix p, and the relationship is precisely this.*0570

*Let us just do an example here. Okay. We will let a equal the matrix (1,2,3)... (0,1,0)... (2,1,2).*0577

*I am not going to go through the entire process, again I use mathematical software to do this... to find the Eigenvalues and to find the Eigenvectors.*0594

*Here is how it works out. As it turns out, one of the Eigenvalues is 4. 4 generates the following Eigenvector, when I do solve the homogenous system, I get (1,0,1).*0602

*A second Eigenvalue is -1. Also real. It generates the Eigenvalue... the Eigenvector (3,0,2). The third Eigenvalue is equal to 1, all distinct, all real, and it generates 1 - 6 and 4, when I solve the homogeneous system.*0619

*Therefore, p = (1,0,1), (-3,0,2), (1,-6,4).*0651

*If I want to define p inverse, which I can, it is not a problem -- you know what I will go ahead and write it out here... it is not going to be too big of an issue -- I have (2/5, -1/5, 0), (7/15, 1/10, -1/6), (3/5, -1/5, 0), and of course my diagonal matrix d is going to end up being (4,0,0), (0,-1,0), (0,0,1).*0667

*If I were to confirm... yes, I would find out that d does in fact equal p inverse × a × p. Excuse me... found the Eigenvalues... found the associated Eigenvector...put those Eigenvectors... these are linearly independent, put them as columns in p... if I take p, multiply and find p inverse, if I multiply a by p inverse on the left, p on the right, I end up with a diagonal matrix where the entries on the main diagonal are precisely the Eigenvalues (4,-1,1), (4,-1,1).*0714

*So, 4 is the first, that is why its Eigenvector is the first column, that is what we meant by respectively.*0753

*Okay. Now, if all the roots of the characteristic polynomial, which is what we solve to find the Eigenvalues, of a are real and distinct... if the roots of the characteristic polynomial are real and distinct... in other words if the Eigenvalues of the matrix are real and distinct, then a is diagonalizable. Always.*0761

*Note how we wrote this. If all of the roots of the characteristic polynomial are real and distinct, then a is diagonalizable. There is no if and only if here.*0812

*That does not mean that if a is diagonalizable, that the roots of the characteristic polynomial, the Eigenvalues are real and distinct.*0822

*It is possible for a to be diagonalizable and have roots that are not distinct. You might have an Eigenvalue, you might have a 4 by 4, and you might have an Eigenvalue 4, and then (1,1,1). That 1 might be an Eigenvalue 3 times over, but it will still be diagonalizable.*0829

*So, again. If then does not mean that it works the same backwards. It is not the same. It is not the same as if and only... if and only if means it goes both ways.*0846

*So, if the roots of the polynomial are real and distinct. If it is diagonalizable, the Eigenvalues may or may not be real and distinct.*0855

*Okay. Now, the characteristic polynomial... so the characteristic poly for non-distinct Eigenvalues looks like this.*0866

*Well, we know we are dealing with some characteristic polynomial... Λ ^{3} + something Λ^{2}, lambda;^{n}.*0890

*Well, every time we find a value, we can factor... that is what the fundamental theorem of algebra says... every polynomial can be factored into linear factors.*0899

*For the non-distinct, we end up with something like this... Λ - Λ _{i} to the k_{i} power.*0910

*So, if some root ends up showing up 5 times, that means that I have 5 factors for that root. Λ - 1, Λ - 1, Λ - 1, Λ - 1, Λ - 1, well, that factor is Λ - Λ _{i}... you know, to this power. Okay.*0917

*This is Λ1, this is Λ 2, in the case of 2, in the case of i... Okay.*0939

*Well, this k _{i} is called multiplicity of the Eigenvalue... Λ_{i}*0974

*It can be shown that if the Eigenvalues of a are all real and distinct, in this case we already dealt with the fact they are distinct, we know they are diagonalizable if they are all real and distinct.*0977

*So, we can also show, now that we have introduced this idea of multiplicity... so if our characteristic polynomial has multiple roots, so if I have a fourth degree equation, one of the roots is 4, the other root is 3, and the other roots are 2 and 2.*1019

*Well, 4 has a multiplicity 1, 3 has a multiplicity 1, 2, because it shows up twice, has a multiplicity 2.*1032

*It can be shown that if the Eigenvalues of a are all real, then a can be diagonalized, it can still be diagonalized if and only if for each Λ _{i} of multiplicity k_{i}, we can find... if we can find k_{i} linearly independent Eigenvectors.*1038

*This means that the null space for that thing that we solved, L _{i}, i_{n} - a, × x = 0, this means that the null space of that equation that we solved to find the Eigenvectors has dimension k_{i}.*1088

*In other words, if I have a root of the characteristic polynomial, an Eigenvalue that has a multiplicity 3... let us say its 1 shows up 3 times.*1118

*Well, when I put it into the equation, this homogeneous equation, if I can actually find 3 linearly independent vectors... if I can find that the dimension of that null space is 3, I can diagonalize that matrix.*1128

*If not, I cannot diagonalize that matrix. Okay.*1145

*So, let us see what we have got. You will let a = (0,0,1), (0,1,2), (0,0,1).*1150

*That is interesting... okay.*1169

*When we take the characteristic polynomial of this, we end up with the following. Λ × (Λ - 1) ^{2}... so we have λ1 = 0, Λ2 = 1, so one has a multiplicity of 2, and Λ3 is also equal to 1.*1173

*Well, let us just deal with this multiplicity of 1 Eigenvalue. When we solved the homogeneous system, we end up with the following. We end up finding that the Eigenvector x is this... (0,r,0).*1195

*Well, this is only 1 vector. Here we have 2 Eigenvalues, multiplicity is 2. Well, in order for this matrix to be diagonalizable, I have to... when I solve that homogeneous system to find the actual Eigenvectors, I need 2 vectors, not just the 1. This is only one vector so this is not diagonalizable.*1214

*Now, mind you, it still has Eigenvalues, and it still has an associated Eigenvector, but it is not diagonalizable. I cannot find some matrix that satisfies that other property.*1240

*Okay. So, now, let us try this one. Let us let a equal to (0,0,0), (0,1,0), (1,0,1).*1255

*Well, as it turns out, this characteristic polynomial is also Λ × (Λ - 1) ^{2}.*1272

*So, again we have Λ = 0, Λ2 = 1, Λ3 = 1, so our Eigenvalue 1 has a multiplicity of 2, so we want to find 2 Eigenvectors.*1280

*We want to find the dimension of this null space associated with this Eigenvalue if it has a dimension 2, we are good. We can actually diagonalize this.*1294

*Let us see what we have. Let us see. When we solve the associated homogeneous system, we end up with (1,0,0,0), (0,0,0,0), (-1,0,0,0).*1302

*When we subject this to reduced row echelon form, it becomes (1,0,0,0), (0,0,0), (0,0,0,0)... so we have... there we go... we have x1 = 0, we have x2 = let us see, this is x2, x3, 3 parameters equals r... x3 = s, and we can rewrite this as equal to r × (0,1,0) + s × (0,0,1), so there you go.*1326

*We have two Eigenvectors, a basis... our basis for that null space has 2 Eigenvectors. It is of dimension 2. It matches the multiplicity of the Eigenvalue, therefore this can be diagonalized.*1375

*So, let us go ahead and actually finish the diagonalization process... when I go back and solve for the Eigenvalue Λ1 = 0, for that Eigenvalue, I get the following vector... r0 - r, as a general, and its specific would be -- let us see -- (1,0,-1).*1392

*Therefore our matrix p would be (1,0,-1), (0,1,0), (0,0,1). This is our matrix p.*1417

*It is of course diagonalizable. Matrix d is going to end up being (0,0,0), (0,1,0), (0,0,1). These along the main diagonal are the Eigenvalues of our matrix. This is our matrix p. We can find the inverse for it, and when we multiply we will find that d, in fact, equals p inverse, a p.*1428

*So, now, procedure for diagonalizing matrix a, this is going to be our recap. First thing we want to do is form the characteristic polynomial, which is symbolized also with f(Λ) equals the determinant of Λ × i _{n}... Λ is our variable... Λ times the identity matrix minus a.*1458

*That is our polynomial. Then, we find the roots of the characteristic polynomial... okay?*1502

*If not all real, if they are not all real, then you can stop... it cannot be diagonalized.*1517

*Okay. Three. For each Eigenvalue, Λ _{i} of multiplicity k_{i}... find a basis for the null space of Λ_{i}, i_{n} - a × x = 0.*1537

*So, for each Eigenvalue Λ _{i} of multiplicity k_{i}, find a basis for the null space... also called the Eigenspace.*1581

*If the dimension... the dimension that we just found... the dimension of this null space that we just found is less than the k _{i}, then you can stop. a cannot be diagonalized.*1594

*If so, well, then, let p be the matrix whose columns are the n linearly independent Eigenvectors... found above... then, the inverse a × p = d, where d has Λ _{i} along the diagonals. The diagonal entries are the Eigenvalues.*1622

*Okay. So, let us recap. We want to form the characteristic polynomial once we are given a matrix a, n by n.*1697

*Once we find the characteristic polynomial, we want to find its roots. If the roots are not all real, if there is... they have to all be real.*1704

*If they are not all real, then you can stop. You cannot diagonalize the matrix.*1713

*If they are real, well, for each Eigenvalue of multiplicity... you know k _{i}, for each one that has a multiplicity, you want to find a basis for the null space of that equation... we solve the homogeneous system.*1718

*Well, if the dimension of that null space is equal to k _{i}, we can continue. If not, we can stop, we cannot diagonalize the matrix.*1735

*But, if we can, and for each distinct Eigenvalue, we are going to have 1 Eigenvector. We will let... take those n linearly independent Eigenvectors that we just found, we will arrange them as columns respectively.*1742

*So Λ1, column 1, Λ2, column 2... that is going to be our matrix p. When we take p inverse × a × p, that actually is equal to our diagonal matrix d, the entries of which are the respective Eigenvalues of our original matrix a.*1757

*So, we will be dealing more with Eigenvalues and Eigenvectors in our next lesson, so we are not quite finished with this, it gets a little bit deeper. We will have a little more exposure with it.*1780

*So, until then, thank you for joining us at Educator.com, and we will see you next time.*1791

2 answers

Last reply by: Hen McGibbons

Sun Apr 24, 2016 5:54 PM

Post by Ahmed Alzayer on September 28, 2015

I have B 2x2 matrix =

Cos x -Sin x

Sin x. Cos x

It still can be diagonalized even though the roots are imaginary, can u clarify.

1 answer

Last reply by: Professor Hovasapian

Wed Nov 13, 2013 3:05 AM

Post by Eddie Chan on November 12, 2013

Hi Raffi,

I received a question about "If A and B are diagonalisable n x n matrix, so is A + B." I have no idea how to prove or disprove it.

0 answers

Post by Manfred Berger on June 23, 2013

In Theorem 2 I get why the multiplicity of the eigenvalues impacts whether or not a matrix is diagonalizable, but why does it matter that the values are real. If the characteristic polynomial has complex roots P has complex entries. Somehow I don't see why that changes anything

0 answers

Post by Manfred Berger on June 23, 2013

Could 2 eigenvectors of an n*n matrix ever be linearly dependent?

1 answer

Last reply by: Carlo Lam

Tue Apr 30, 2013 12:54 AM

Post by Carlo Lam on April 30, 2013

P is always an arbitrary matrix?

0 answers

Post by Matt C on April 27, 2013

I guess I don't understand what you are saying for example 4 at 20:18 when you say this is only one vector. When I plugged in all the values for lambda, I got three eigenvectors [[1,0,0], [0,1,0], [0,1,0]]. If there is a way where you could explain it would be nice. I was feeling pretty good up until this point with eigenvalues, eigenvectors, and diagonalization.